Discussion
An aeroplane's Dutch roll damping ratio improves from 0.06 to 0.18 when the yaw damper is engaged, the frequency staying at \( 1.4 \ \mathrm{rad/s} \). The time to half amplitude changes from
Answer: C
The decay time is \( \ln 2/(\zeta\omega_n) \) in each case:
\[ \begin{aligned} t_{1} &= \frac{0.693}{0.06 \times 1.4} = 8.3 \ \mathrm{s} \\ t_{2} &= \frac{0.693}{0.18 \times 1.4} = 2.8 \ \mathrm{s} \end{aligned} \]
so 8.3 s to 2.8 s, a threefold improvement matching the threefold rise in damping ratio, since the frequency is unchanged. The second option scales only part way. The third and fourth use a different frequency. With frequency fixed, decay time is simply inversely proportional to the damping ratio.
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